KCET · Physics · Gravitation
What is the value of acceleration due to gravity at a height equal to half the radius of the Earth, from its surface?
- A \(4.4 \mathrm{~ms}^{-2}\)
- B \(6.5 \mathrm{~ms}^{-2}\)
- C Zero
- D \(9.8 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(A) \(4.4 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Acceleration due to gravity at height \(h\) from the surface of the earth.
\(g_h=\frac{g}{\left(1+\frac{h}{R}\right)^2}\)
Given, \(\quad h=\frac{R}{2}\)
\(\therefore \quad g_h=\frac{g}{\left(1+\frac{R}{R}\right)^2}=\frac{g}{\left(\frac{3}{2}\right)^2}\)
\(=\frac{4}{9} g=\frac{4}{9} \times 9.8=4.4 \mathrm{~m} / \mathrm{s}^2\)
\(g_h=\frac{g}{\left(1+\frac{h}{R}\right)^2}\)
Given, \(\quad h=\frac{R}{2}\)
\(\therefore \quad g_h=\frac{g}{\left(1+\frac{R}{R}\right)^2}=\frac{g}{\left(\frac{3}{2}\right)^2}\)
\(=\frac{4}{9} g=\frac{4}{9} \times 9.8=4.4 \mathrm{~m} / \mathrm{s}^2\)
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